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(1 + i)⁶ + (1 – i) ³ = {(1 + i)² }³ + (1 – i)² (1 – i) 

= {1 + i² + 2i}³ + (1 + i² – 2i)(1 – i) 

= {1 – 1 + 2i} ³ + (1 – 1 – 2i)(1 – i) 

= (2i)³ + (– 2i)(1 – i) 

= 8i³ + (– 2i) + 2i² 

[since i³ = – i, i² = – 1] 

= – 8i – 2i – 2 

= – 10 i – 2 

= – 2(1 + 5i)

1+ i² + i⁴ + i⁶ + i⁸ + ... + i²⁰

= 1 + (– 1) + 1 + (– 1) + 1 + ... + 1 

= 1

i⁵ + i¹⁰ + i¹⁵ = i ^{(4 + 1)} + i ^{(8 + 2)} + i ^{(12 + 3)} 

= (i⁴)¹×i + (i⁴)²×i² + (i⁴)³×i³ 

= (i⁴)¹×i + (i⁴)²×i² + (i⁴)³×i²×i 

= 1×i + 1×(– 1) + 1×(– 1)×i 

= i – 1 – i = – 1

i + i² + i³ + i⁴ = i + i² + i²×i + i⁴ 

= i – 1 + (– 1)×i + 1 

since i⁴ = 1, i² = – 1 

= i – 1 – i + 1 = 0

i³⁰ + i⁸⁰ + i¹²⁰ = i^{(28 + 2)} + i⁸⁰ + i¹²⁰ 

= (i⁴)⁷ × i² + (i⁴)²⁰ + (i⁴)³⁰ 

= – 1 + 1 + 1 = 1

[since i⁴ = 1, i² = – 1] 

i³⁰ + i⁸⁰ + i¹²⁰ = 1

i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = i^{(48 + 1)} + i⁶⁸ + i^{(88 + 1}) + i^{(108 + 2)} 

= (i⁴)¹² × i + (i⁴)¹⁷ + (i⁴)¹¹ × i + (i⁴)²⁷ × i²

= i + 1 + i – 1 = 2i 

[since i⁴ = 1, i² = – 1] 

i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = 2i

1 + i¹⁰ + i²⁰ + i³⁰ = 1 + i ^{(8 + 2)} + i²⁰ + i ^{(28 + 2}) 

= 1 + (i⁴)² × i² + (i⁴)⁵ + (i⁴)⁷ × i² 

= 1 – 1 + 1 – 1 = 0 

[since i⁴ = 1, i² = – 1] 

Hence , 1 + i¹⁰ + i²⁰ + i³⁰ is a real number.

z = 2 + 3i

∴ \(\bar z\) = 2 – 3i

∴ \(\bar z\) = 2 + 3i = z

z \(\bar z\)= (2 + 3i) (2 – 3i)

= 4 – 9i²

= 4 – 9(-1) …..[∵ i = -1]

= 13

|z|² = \((\sqrt{2^2+3^2})^2\) 

= 2² + 3²

= 4 + 9

 = 13
\(\overline{z\bar z}\) = |z|²

z – \(\bar z\)= (2 + 3i) – (2 – 3i)

= 2 + 3i – 2 + 3i

= 6i

abi = 3a – b + 12i

∴ 0 + abi = (3a – b) + 12i

Equating real and imaginary parts, we get

3a – b = 0

∴ 3a = b …..(i)

and ab = 12

∴ b = 12/a……..(ii)

Substituting b = 12/a in (i), we get

3a = 12/a 

3a²= 12

a² = 4

a = ±2

When a = 2, b = 12/a = 12/2 = 6

When a = -2, b = 12/a = 12/-2 = -6

∴ a = 2 and b = 6 or a = -2 and b = -6

(z + \(\bar z\)) = (2 + 3i) + (2 – 3i)

= 2 + 3i + 2 – 3i

= 4, which is a real number.

∴ z + \(\bar z\) is real.

∣z + 4∣ ≤ 3 ⇒−3 ≤ z + 4 ≤ 3

⇒−6 ≤ z + 1 ≤ 0

⇒0 ≤ −(z + 1) ≤ 6

⇒0 ≤ ∣z + 1∣ ≤ 6 

Hence, the greatest and least values are 6 and 0.

we have, 3i³ - 2ai² + (1 – a)i+ 5 

⟹ -3i+ 2a + (1 – a)i + 5 

⟹ (2a + 5) + i(1 – a – 3) 

⟹ (2a + 5) – i(2 + a)

(a + ib)(1 + i) = 2 + i

a + ai + bi + bi² = 2 + i

a + (a + b)i + b(-1) = 2 + i ……(∵ i² = -1)

(a – b) + (a + b)i = 2 + i

Equating real and imaginary parts, we get

a – b = 2 ……(i)

a + b = 1 …….(ii)

Adding equations (i) and (ii),

we get

2a = 3

∴ a = 3/2

Substituting a = 3/2 in (ii), we get

3/2 + b = 1

∴ b = 1 - 3/2 = -1/2

∴ a = 3/2 and b = -1/2

a + 2b + 2ai = 4 + 6i

Equating real and imaginary parts, we get

a + 2b = 4 …..(i)

2a = 6 ……(ii)

∴ a = 3

Substituting, a = 3 in (i), we get

3 + 2b = 4

∴ b = 1/2

∴ a = 3 and b = 1/2

Check:

For a = 3 and b = 1/2

Consider, L.H.S. = a + 2b + 2ai

= 3 + 2(1/2) + 2(3)i

= 4 + 6i

= R.H.S.

(a + b) (2 + i) = b + 1 + (10 + 2a)i

2(a + b) + (a + b)i = (b + 1) + (10 + 2a)i

Equating real and imaginary parts, we get

2(a + b) = b + 1

∴ 2a + b = 1 ……(i)

and a + b = 10 + 2a

-a + b = 10 …….(ii)

Subtracting equation (ii) from (i), we get

3a = -9 

∴ a = -3

Substituting a = – 3 in (ii), we get

-(-3) + b = 10

∴ b = 7

∴ a = -3 and b = 7

(a – b) + (a + b)i = a + 5i

Equating real and imaginary parts, we get

a – b = a ……(i)

a + b = 5 ……(ii)

From (i), b = 0

Substituting b = 0 in (ii),

we get a + 0 = 5

∴ a = 5

∴ a = 5 and b = 0

z = a + ib 

|a+ib-5i| = |a+ib+5i| 

|a+ib-5i|² = |a+ib+5i|² 

|a +i(b-5)|² = |a + i(b+5)|² 

a²+(b-5)² = a²+(b+5)² 

a²+b²+25-10b = a²+b²+25+10b 

20b = 0 

b = 0 

b is a imaginary part of z

Im(z) = \(\frac{z-\bar z}{2}\)

= y 

= 0 

Im (z) = 0 

So, the locus point is real axis

\(\sqrt{-25} \times \sqrt{-9}\) = \(\sqrt{25}\sqrt{-1}\times\sqrt{9}\sqrt{-1}\)

= 5i×3i 

= 15i² 

= -15

(i)  (5 + 9 i) (-3 + 4 i)

= 5(-3 + 4 i ) +9 i(-3 + 4 i)

= -15 + 20 i -27 i +36 i²

= -15 - 7 i - 36

= -51 - 7 i

= x = -51 y = -7

(II) (-2 - 5 i) (3 - 6 i)

 = -2(3 - 6 i) -5 i(3 - 6 i)

= -6 + 12 i - 15 i + 30 i²

= -6 -3 i +30

= 24 - 3 i

= x = 24 y = -3

As we all know that,

z = r(cos θ + i sin θ ) , θ = arg(z) 

\((\bar z)\)  = r(cos(θ) - sin(θ) 

= r(cos(-θ)+sin(-θ)

-θ = arg \((\bar z)\)    

So, arg(z) + arg\((\bar z)\) = θ - θ

= 0

Given complex number is Z=1-i 

We know that the multiplicative inverse of a complex number Z is Z^-1 (or) \(\frac{1}{z}\)

⇒ Z^-1 = \(\frac{1}{1-i}\)

Multiplying and dividing with 1+i

⇒ Z^-1 = \(\frac{1}{1-i}\times \frac{1+i}{1+i}\) 

⇒ Z^-1 = \(\frac{1+i}{1^2-(i)^2}\) 

We know that i²=-1

⇒ Z^-1 = \(\frac{1+i}{1-(-1)}\) 

⇒ Z^-1 = \(\frac{1+i}{2}\) 

∴ The Multiplicative inverse of 1-i is \(\frac{1+i}{2}\)

Given complex number is Z= √5+3i

We know that the multiplicative inverse of a complex number Z is Z^-1 (or) \(\frac{1}{Z}\).

⇒ Z^-1 =  \(\frac{1}{√{5} + 3i}\)

Multiplying and dividing with  √5+3i

⇒ Z^-1  =\(\frac{1}{√5+3i}\times\frac{√5- 3i}{√5- 3i}\)

⇒ Z^-1  = \(\frac{√5 - 3i}{(√5 )^2-(3i)^2}\)

⇒ Z^-1  =  \(\frac{√5+3i}{5-9i^2}\)

We know that i²=-1

⇒ Z^-1  = \(\frac{√5+3i}{5-9(-1)}\)

⇒ Z^-1  = \(\frac{√5+3i}{14}\)

∴ The Multiplicative inverse of √5+3i is \(\frac{√5+3i}{14}\)

Given that (1 + i) (1 + 2i) (1 + 3i) = x + i y …(1) 

We can also say that

(1 - i) (1 - 2i) (1 - 3i) = x - i y …(2) 

Multiply and divide the eq no. 2 with eq no. 1

\(\frac{ (1 + i) (1 - i)(1 + 2i)(1 - 2i)(1 + 3i)(1 - 3i)}{(1 - i) (1 - 2i) (1 - 3i)}\) = \(\frac{(x + i y)(x - i y)}{x - i y}\) 

((1)² – (i)²)((1)² – (2i)²)((1)² – (3i)²) = ((x)² – (iy)²) 

x² + y² = 2 × 5 × 10 = 100

|z| = \(\sqrt{(1-cosθ)^2 + (sinθ)^2}\) 

=\(\sqrt{2-2cosθ}\) 

=  \(2|sin\frac{θ}{2}|\) 

\(\frac{1+a}{1-a}\) = \(\frac{1+(cosθ + isinθ)}{1-(cosθ + isinθ)}\) = \(\frac{1+(cosθ + isinθ)}{1-(cosθ - isinθ)}\times \frac{(1-(cosθ )+ isinθ}{(1-cosθ )+ isinθ}\)

= \(\frac{2isinθ}{2-2cosθ}\) 

= 0 + \(i\frac{sinθ}{1-cosθ}\) 

= \(icot\frac{θ}{2}\) 

False

Explanation:

Let z = x + iy, where x, y > 0

i.e. z or point A (x, y) lies in first quadrant.

Now, -iz = -I (x + iy)

= -ix – i²y

= y – ix

Now, point B (y, -x) lies in fourth quadrant.

Also ∠AOB = 90°

Thus, B is obtained by rotating A in clockwise direction about origin.

Given as

Z = sin π/5 + i(1 – cos π/5)

On using the formula,

sin 2θ = 2 sin θ cos θ

1 - cos 2θ = 2 sin² θ

Therefore,

Z = 2 sin π/10 cos π/10 + i(2 sin² π/10)

= 2 sin π/10 (cos π/10 + i sin π/10)

Thus, the polar form of sin π/5 + i(1 – cos π/5) is 2 sin π/10(cos π/10 + i sin π/10)

(-i)^{4n + 1} 

= (i². i)^{4n + 1} 

= (i³)^{4n + 1}

= i^{12n + 3}

= i¹²ⁿ. i³

= (i⁴)³ⁿ. i². i

= 1. (-1). i

= -i.

Given: 

⇒ z²+|z|²=0 

Let us assume z=x+iy 

⇒ (x+iy)² + \(\sqrt{(x^2+y^2)^2}\) = 0

⇒ x²+(iy)²+2(x)(iy)+x²+y²=0 

⇒ 2x²+y²+i²y²+i²xy=0 

We know that i²=-1 

⇒ 2x²+y²-y²+i²xy=0 

⇒ 2x²+i²xy=0 

Equating Real and Imaginary parts on both sides we get, 

⇒ 2x²=0 and 2xy=0 

⇒ x=0 and y\(\varepsilon\)R 

∴ z=0+iy where y\(\varepsilon\)R. i.e, Infinite solutions.

 \(\frac{x}{1+i}+\frac{y}{2-i}=\frac{1-5i}{3-2i}\) 

⇒ \(\frac{x}{1+i}\times\frac{1-i}{1-i}+\frac{y}{2-i}\times\frac{2+i}{2+i}\) = \(\frac{1-5i}{3-2i}\times\frac{3+2i}{3+2i}\)

⇒\(\frac{x-xi}{1-i^2}+\frac{2y+yi}{4-i^2}\) = \(\frac{3-15i+2i-10i^2}{9-4i^2}\) 

(∵ (a + b)(a - b) = a² - b²)

⇒\(\frac{1}{2}\) (x - xi) + \(\frac{1}{5}\)(2y + yi) = \(\frac{1}{13}\) (13 - 13i) (∵i² = -i)

⇒ \(\Big(\frac{x}{2}+\frac{2y}{5}\Big)\) + i\(\Big(\frac{-x}{2}+\frac{y}{5}\Big)\) = 1 - i

⇒ \(\frac{x}{2}+\frac{2y}{5}\) = 1 and \(\frac{-x}{2}+\frac{y}{5}\) = -1  (By comparing real & imaginary part of complex numbers)

⇒ \(\frac{5y}{5}\) = 1 - 1 = 0 (By adding both equations)

⇒ y = 0

Put y = 0 in \(\frac{x}{2}+\frac{2y}{5}\)= 1, we get

\(\frac{x}{2}\) = 1 

⇒ x = 2

Hence, solution of given complex equation is x = 2, y = 0

(A) -3√2

√-3 √-6 

= (√3 i) (√6 i)

= 3√2 (-1)

= -3√2

Given as

|z₁| = |z₂| and arg (z₁) + arg (z₂) = π

Let us assume that arg (z₁) = θ

arg (z₂) = π – θ

As we know that in the polar form, z = |z| (cos θ + i sin θ)

z₁ = |z₁| (cos θ + i sin θ) …………. (i)

z₂ = |z₂| (cos (π – θ) + i sin (π – θ))

= |z₂| (-cos θ + i sin θ)

= – |z₂| (cos θ – i sin θ)

Then let us find the conjugate of

bar z₂ = – |z₂| (cos θ + i sin θ) …… (ii) (since, |bar z₂ = |z₂|)

Then,

z₁/bar z₂ = [|z₁| (cos θ + i sin θ)]/[-|z₂| (cos θ + i sin θ)]

= – |z₁|/|z₂| [since, |z₁| = |z₂|]

= -1

Then, when we cross multiply we get,

z₁ = – bar z₂
Thus proved.

(D) z lies on a circle

|z – zi | = |z| |1 – i| = 1

\(\therefore\) |z| = 1/√2

\(\therefore\) x² + y² = 1/2

Let z = -i = 0 – i

a = 0, b = -1

z lies on negative imaginary Y-axis.

|z| = r = \(\sqrt{a^2+b^2}=\sqrt{0^2+(-1)^2}=1\) and 

θ = amp z = 270° = \(\frac{3\pi}2\) 

 The polar form of z = r (cos θ + i sin θ)

= 1 (cos 270° + i sin 270°)

= 1\((cos\frac{3\pi}2+i sin\frac{3\pi}2)\)

The exponential form of z = re^iθ = e^{\(\frac{3\pi}2i\)}

Conjugate of (√(-5) – √7 i) is (√(-5) + √7i).

Answer: 1

p + iq = (a + ib/a - ib) then p² + q² = 1

Conjugate of (5i) is (-5i).

-√(-5) = -√5 × √(-1) = -√5 i

Conjugate of (-√(-5)) is √5 i

\(\frac{2i}{1+i}\) = \(\frac{2i}{(1+i)}\times\frac{(1-i}{1-i}\) = 1+i

\(\Big(\frac{2i}{1+i}\Big)^n\) ⁿ= (1+i)

Let check the value of (1 + i)ⁿ for different value of n 

at n =1 , 1+ i (no) 

at n =2 , (1 + i)² = 1 + i² + 2i = 2i (no) 

at n =3 , (1 + i)²(1 + i) = (1 + i)(2i) = 2i – 2 (no) 

at n =4 , (1 + i)²(1 + i)² = (2i)² = -4 (no) 

at n =5 , (1 + i)⁴(1 + i) = -4(1 + i) (no) 

at n =6 , (1 + i)⁴(1 + i)² = -4(2i) (no) 

at n =7 , (1 + i)⁶(1 + i) = -8i(1 + i) = -8i + 8 (no) 

at n =8 , (1 + i)⁴(1 + i)⁴ = (-4)(-4) = 8 (yes) 

So, we can say that n = 8 is the least positive integer for which \(\Big(\frac{2i}{1+i}\Big)^n\) is positive integer

Let, (a + ib)² = 3 + 4√7 i

Now using, (a + b)² = a² + b² + 2ab

⇒ a² + (bi)² + 2abi = 3 + 4√7 i

Since i² = -1

⇒ a² - b² + 2abi = 3 + 4√7 i

Now, separating real and complex parts, we get

⇒ a² - b² = 3 …………..eq.1

⇒ 2ab = 4√7 …….. eq.2

⇒ a = 2√7/b

Now, using the value of a in eq.1, we get

⇒ \((\frac{2\sqrt7}{b})^2\) – b² = 3

⇒ 12 – b⁴ = 3b²

⇒ b⁴ + 3b² - 28 = 0

Simplify and get the value of b², we get,

⇒ - b² = -7 or b² = 4

as b is real no. so, b² = 4

b = 2 or b = - 2

Therefore, a = √7 or a = -√7

Hence the square root of the complex no. is √7 + 2i and -√7 - 2i.

Given: z = √2

The above can be re – written as

z = √2 + 0i

Here, the imaginary part is zero

So, the conjugate of z = √2 + 0i is

\(\bar z\) = √2 - 0i

Or \(\bar z\) = √2

Conjugate of (3 + i) is (3 – i).

Given: z = (2 – 5i)²

First we calculate (2 – 5i)² and then we find the conjugate

(2 – 5i)² = (2)² + (5i)² – 2(2)(5i) = 4 + 25i² – 20i

= 4 + 25(-1) – 20i [∵ i² = -1]

= 4 – 25 – 20i = -21 – 20i

Now, we have to find the conjugate of (-21 – 20i)

So, the conjugate of (- 21 – 20i) is (-21 + 20i)

x + iy)^1/3 = a + ib 

x + iy = (a + ib)³ 

= a³ + (ib)³ + 3a²(ib) + 3a(ib)² 

= a³ – ib³ + i3a²b – 3ab² 

= (a³ – 3ab²) + i(3a²b – b3) 

x = a³ – 3ab² and y = 3a²b – b³

\(\frac{x}{a}+\frac{y}{b}\)  = \(\frac{a^3-3ab^2}{a}+\frac{3a^2b-b^3}{b}\)

= a³ – 3b² + 3a² – b³

= 4(a² – b²)

\(\sqrt{a+ib}\) = x+iy

Square both sides 

a + ib = (x + iy)² = x² + i2xy - y² 

So, we can say that a = x² – y² and b = 2xy 

a – ib = (x² – y²) – i(2xy) 

= (x)² + 2(x)(-iy) + (-iy)² 

= (x + (-iy))² 

= (x – iy)²

\(\sqrt{a-ib}\) = x-iy

\(4\sqrt{-4}+5\sqrt{-9}-3\sqrt{-16}\) 

= \(4\sqrt{4\times-1}+5\sqrt{9\times-1}-3\sqrt{16\times-1}\) 

= 4(2i) + 5(3i) - 3(4i)

= 8i + 15i - 12i

= 11i

\(\sqrt{-16}+3\sqrt{-25}+\sqrt{-36}-\sqrt{-625}\)

= \(\sqrt{16\times-1}+3\sqrt{25\times-1}+\sqrt{36\times-1}-\sqrt{625\times-1}\) 

= 4i + 3(5i) + 6i - 25i

= 25i - 25i

= 0

Since i = \((\sqrt{-1})\) so

(i) L.H.S. = \((\sqrt{-1})^{192}\)

⇒ i¹⁹²

⇒ i^{4 x 48} = 1

Since it is of the form i⁴ⁿ = 1 so the solution would be 1

(ii) L.H.S.= \((\sqrt{-1})^{93}\)

⇒ i^{4 x 23 + 1}

⇒ i^{4n + 1}

⇒ i¹ = i

Since it is of the form of i^{4n + 1} = i so the solution would be simply i.

(iii) L.H.S = \((\sqrt{-1})^{30}\)

⇒ i^{4 x 7 + 2}

⇒ i^{4n + 2}

⇒ i² = -1

Since it is of the form i^{4n + 2} so the solution would be -1